# 给你一个链表，两两交换其中相邻的节点，并返回交换后链表的头节点。你必须在不修改节点内部的值的情况下完成本题（即，只能进行节点交换）。
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#  示例 1：
# 输入：head = [1,2,3,4]
# 输出：[2,1,4,3]
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#  示例 2：
# 输入：head = []
# 输出：[]
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#  示例 3：

# 输入：head = [1]
# 输出：[1]
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#  Related Topics 递归 链表 👍 2194 👎 0

from com.example.linked.common import *


class Solution:
    def swapPairs1(self, head: Optional[ListNode]) -> Optional[ListNode]:
        """
        迭代实现
        """
        dummy_head = ListNode(-1)
        dummy_head.next = head
        temp = dummy_head
        while temp.next and temp.next.next:
            node1 = temp.next
            node2 = temp.next.next
            temp.next = node2
            node1.next = node2.next
            node2.next = node1
            temp = node1
        return dummy_head.next

    def swapPairs2(self, head: Optional[ListNode]) -> Optional[ListNode]:
        """
        递归实现
        """
        if not head or not head.next:
            return head
        new_head = head.next
        head.next = self.swapPairs2(new_head.next)
        new_head.next = head
        return new_head

    def swapPairs(self, head: Optional[ListNode]) -> Optional[ListNode]:
        return self.swapPairs1(head)


if __name__ == '__main__':
    head = getListNode(1, 2, 3, 4)
    res = Solution().swapPairs(head)
    vistListNode(res)
